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AP Chemistry

AP Chemistry Help: Moles And Molar Mass

Review real example questions for Moles And Molar Mass in AP Chemistry.

Question 1

A sample of pure aluminum, Al\mathrm{Al}Al, contains 3.01×10233.01\times 10^{23}3.01×1023 atoms. How many moles of Al\mathrm{Al}Al are in the sample? (Avogadro's number: 6.022×1023 mol−16.022\times 10^{23}\ \mathrm{mol^{-1}}6.022×1023 mol−1.)

  1. 0.250 mol0.250\ \mathrm{mol}0.250 mol
  2. 2.00 mol2.00\ \mathrm{mol}2.00 mol
  3. 6.02×1023 mol6.02\times 10^{23}\ \mathrm{mol}6.02×1023 mol
  4. 0.500 mol0.500\ \mathrm{mol}0.500 mol
  5. 1.00 mol1.00\ \mathrm{mol}1.00 mol
Explanation: This question tests the skill of moles and molar mass. Molar mass relates mass to moles, but this problem involves converting from particles to moles using Avogadro's number. Avogadro's number bridges moles to the number of atoms or molecules, allowing us to find moles by dividing the number of particles by 6.022 × 10^23 per mole. For elemental aluminum, this calculation reveals the amount in moles from a count of atoms. A common distractor is choice C, 6.02 × 10^23 mol, which comes from mistakenly multiplying instead of dividing by Avogadro's number, inverting the conversion factor. Ensure you divide when going from particles to moles. A transferable strategy is to write the unit you want (moles) and use Avogadro's number as the bridge by dividing particles by particles per mole.

Question 2

A sample of pure potassium bromide, KBr\mathrm{KBr}KBr, has a mass of 11.9 g11.9\ \mathrm{g}11.9 g. (Atomic masses: K=39.1\mathrm{K}=39.1K=39.1, Br=79.9\mathrm{Br}=79.9Br=79.9.) How many moles of KBr\mathrm{KBr}KBr are in the sample?

  1. 11.9 mol11.9\ \mathrm{mol}11.9 mol
  2. 0.0500 mol0.0500\ \mathrm{mol}0.0500 mol
  3. 0.200 mol0.200\ \mathrm{mol}0.200 mol
  4. 1.00 mol1.00\ \mathrm{mol}1.00 mol
  5. 0.100 mol0.100\ \mathrm{mol}0.100 mol
Explanation: This question tests the skill of moles and molar mass. Molar mass links mass to amount through moles = mass / molar mass, with KBr's molar mass as 39.1 + 79.9 = 119 g/mol. This allows direct calculation from given mass. Avogadro's number, 6.02 × 10²³, links moles to particles for further conversions if needed. A tempting distractor is choice A, 0.0500 mol, from a common unit-conversion error of using half the molar mass, perhaps treating Br as 40 instead of 79.9. A transferable strategy is to write the unit you want and use molar mass as the bridge.

Question 3

A 23.0 g23.0\ \mathrm{g}23.0 g sample of pure ethanol, C2H6O\mathrm{C_2H_6O}C2​H6​O, is used. (Atomic masses: C=12.0\mathrm{C}=12.0C=12.0, H=1.0\mathrm{H}=1.0H=1.0, O=16.0\mathrm{O}=16.0O=16.0.) How many moles of ethanol are present?

  1. 0.250 mol0.250\ \mathrm{mol}0.250 mol
  2. 0.500 mol0.500\ \mathrm{mol}0.500 mol
  3. 1.00 mol1.00\ \mathrm{mol}1.00 mol
  4. 2.00 mol2.00\ \mathrm{mol}2.00 mol
  5. 46.0 mol46.0\ \mathrm{mol}46.0 mol
Explanation: This question tests the skill of moles and molar mass. Molar mass links mass to amount as moles = mass / molar mass, with ethanol at 46.0 g/mol giving 0.500 mol. This conversion is fundamental. Avogadro's number, 6.02 × 10²³, links moles to particles. A tempting distractor is choice A, 0.250 mol, from a common unit-conversion error of halving the mass. A transferable strategy is to write the unit you want and use molar mass as the bridge.

Question 4

A student measures out 44.0 g44.0\ \text{g}44.0 g of pure carbon dioxide, CO2\text{CO}_2CO2​. (Molar mass of CO2=44.0 g/mol\text{CO}_2=44.0\ \text{g/mol}CO2​=44.0 g/mol.) How many moles of CO2\text{CO}_2CO2​ are present?

  1. 2.00 mol2.00\ \text{mol}2.00 mol
  2. 0.0227 mol0.0227\ \text{mol}0.0227 mol
  3. 44.0 mol44.0\ \text{mol}44.0 mol
  4. 1.00 mol1.00\ \text{mol}1.00 mol
  5. 88.0 mol88.0\ \text{mol}88.0 mol
Explanation: This question tests the skill of moles and molar mass. Molar mass provides the conversion factor between mass and moles through the relationship: moles = mass ÷ molar mass. With 44.0 g of CO₂ and a molar mass of 44.0 g/mol, we calculate: 44.0 g ÷ 44.0 g/mol = 1.00 mol. Avogadro's number would be necessary to find the number of molecules, but here we only need moles. A tempting wrong answer (choice A, 0.0227 mol) results from incorrectly inverting the calculation and dividing 1 by 44.0 instead of dividing 44.0 by 44.0. To solve mole conversions accurately, set up the calculation so that grams cancel out, leaving moles as your final unit.

Question 5

A sample contains 0.500 mol0.500\ \text{mol}0.500 mol of pure magnesium metal, Mg\text{Mg}Mg. How many magnesium atoms are in the sample? (Use NA=6.02×1023 mol−1N_A = 6.02\times10^{23}\ \text{mol}^{-1}NA​=6.02×1023 mol−1.)

  1. 1.20×1023 atoms1.20\times10^{23}\ \text{atoms}1.20×1023 atoms
  2. 3.01×1023 atoms3.01\times10^{23}\ \text{atoms}3.01×1023 atoms
  3. 6.02×1023 atoms6.02\times10^{23}\ \text{atoms}6.02×1023 atoms
  4. 0.500 atoms0.500\ \text{atoms}0.500 atoms
  5. 1.00×1024 atoms1.00\times10^{24}\ \text{atoms}1.00×1024 atoms
Explanation: This question tests the skill of moles and molar mass. While molar mass links mass to moles, Avogadro's number (6.02×10²³ particles/mol) connects moles to the number of particles (atoms, molecules, or formula units). To find atoms from moles, we multiply: 0.500 mol × 6.02×10²³ atoms/mol = 3.01×10²³ atoms. The molar mass of Mg would only be needed if we were converting between mass and moles, but here we're converting between moles and atoms. A common error (choice C, 6.02×10²³ atoms) results from forgetting to multiply by the number of moles and just using Avogadro's number directly. To convert from moles to particles, always multiply the number of moles by Avogadro's number to get the total particle count.

Question 6

A 18.0 g18.0\ \text{g}18.0 g sample of pure glucose, C6H12O6\text{C}_6\text{H}_{12}\text{O}_6C6​H12​O6​, is placed in a container. How many moles of glucose are present? (Molar masses: C=12.0 g/mol\text{C}=12.0\ \text{g/mol}C=12.0 g/mol, H=1.0 g/mol\text{H}=1.0\ \text{g/mol}H=1.0 g/mol, O=16.0 g/mol\text{O}=16.0\ \text{g/mol}O=16.0 g/mol.)

  1. 0.300 mol0.300\ \text{mol}0.300 mol
  2. 0.100 mol0.100\ \text{mol}0.100 mol
  3. 0.600 mol0.600\ \text{mol}0.600 mol
  4. 1.00 mol1.00\ \text{mol}1.00 mol
  5. 180 mol180\ \text{mol}180 mol
Explanation: This problem tests moles and molar mass by converting mass to moles for a complex molecule. The molar mass of glucose (C₆H₁₂O₆) equals 6 carbon atoms (6 × 12.0 = 72.0 g/mol) plus 12 hydrogen atoms (12 × 1.0 = 12.0 g/mol) plus 6 oxygen atoms (6 × 16.0 = 96.0 g/mol), totaling 180.0 g/mol. To find moles: 18.0 g ÷ 180.0 g/mol = 0.100 mol. A common mistake would be to use 18.0 as the molar mass directly, giving 1.00 mol. The key strategy is to systematically calculate molar mass by counting each type of atom and multiplying by its atomic mass before adding them all together.

Question 7

A student has a 18.0 g18.0\text{ g}18.0 g sample of pure water, H2O\text{H}_2\text{O}H2​O. What amount of H2O\text{H}_2\text{O}H2​O, in moles, is present? (Molar mass of H2O=18.0 g/mol\text{H}_2\text{O}=18.0\text{ g/mol}H2​O=18.0 g/mol.)​​

  1. 0.500 mol0.500\text{ mol}0.500 mol
  2. 18.0 mol18.0\text{ mol}18.0 mol
  3. 36.0 mol36.0\text{ mol}36.0 mol
  4. 1.00 mol1.00\text{ mol}1.00 mol
  5. 0.0556 mol0.0556\text{ mol}0.0556 mol
Explanation: This problem tests the skill of moles and molar mass. Molar mass serves as the conversion factor between mass (in grams) and amount (in moles), allowing us to determine how many moles are in a given mass of substance. To convert from grams to moles, we divide the mass by the molar mass: moles = mass ÷ molar mass. For this water sample: moles = 18.0 g ÷ 18.0 g/mol = 1.00 mol. A common error would be multiplying instead of dividing (18.0 × 18.0 = 324), which doesn't appear here but would give an incorrect large value. When converting mass to moles, always divide by molar mass—think of it as 'how many molar mass units fit into your sample mass.'

Question 8

A student has 0.500 mol0.500\ \mathrm{mol}0.500 mol of pure ammonia, NH3(s)\mathrm{NH_3(s)}NH3​(s). The molar mass of NH3\mathrm{NH_3}NH3​ is 17.0 g mol−117.0\ \mathrm{g\ mol^{-1}}17.0 g mol−1. What is the mass of the ammonia sample?

  1. 34.0 g34.0\ \mathrm{g}34.0 g
  2. 8.50 g8.50\ \mathrm{g}8.50 g
  3. 17.0 g17.0\ \mathrm{g}17.0 g
  4. 0.0294 g0.0294\ \mathrm{g}0.0294 g
  5. 0.500 g0.500\ \mathrm{g}0.500 g
Explanation: This problem requires using moles and molar mass to convert from moles to mass. The molar mass of NH₃ is 17.0 g/mol, which tells us the mass of one mole. To find the mass of 0.500 mol, we multiply: 0.500 mol × 17.0 g/mol = 8.50 g. A student might mistakenly divide the molar mass by moles (17.0 ÷ 0.500 = 34.0), which gives twice the molar mass rather than half. The reliable method is dimensional analysis: start with moles, multiply by molar mass (g/mol), and verify that mol cancels to leave grams.

Question 9

A student measures out 0.100 mol0.100\text{ mol}0.100 mol of pure magnesium chloride, MgCl2\text{MgCl}_2MgCl2​. How many formula units of MgCl2\text{MgCl}_2MgCl2​ does this sample contain? (Use NA=6.02×1023 mol−1N_A=6.02\times10^{23}\text{ mol}^{-1}NA​=6.02×1023 mol−1.)​​

  1. 6.02×10226.02\times10^{22}6.02×1022 formula units
  2. 6.02×10236.02\times10^{23}6.02×1023 formula units
  3. 3.01×10233.01\times10^{23}3.01×1023 formula units
  4. 1.00×10−11.00\times10^{-1}1.00×10−1 formula units
  5. 1.20×10241.20\times10^{24}1.20×1024 formula units
Explanation: This problem involves moles and molar mass concepts. Avogadro's number bridges between moles and particles, with one mole containing 6.02×10²³ formula units regardless of the substance. To find the number of formula units from moles, we multiply: formula units = moles × NA. For this MgCl₂ sample: formula units = 0.100 mol × 6.02×10²³ mol⁻¹ = 6.02×10²² formula units. A tempting error would be forgetting to account for the 0.100 coefficient and selecting 6.02×10²³ (choice B), which represents one full mole. When converting moles to particles, multiply by Avogadro's number and carefully track your decimal places or powers of ten.

Question 10

A sample contains 0.100 mol0.100\ \mathrm{mol}0.100 mol of pure aluminum metal, Al(s)\mathrm{Al(s)}Al(s). The molar mass of aluminum is 27.0 g mol−127.0\ \mathrm{g\ mol^{-1}}27.0 g mol−1. What is the mass of the aluminum sample?

  1. 270 g270\ \mathrm{g}270 g
  2. 2.70 g2.70\ \mathrm{g}2.70 g
  3. 0.00370 g0.00370\ \mathrm{g}0.00370 g
  4. 27.0 g27.0\ \mathrm{g}27.0 g
  5. 0.100 g0.100\ \mathrm{g}0.100 g
Explanation: This problem tests converting moles to mass using moles and molar mass. The molar mass of aluminum (27.0 g/mol) tells us that one mole of aluminum has a mass of 27.0 grams. To find the mass of 0.100 mol, we multiply: 0.100 mol × 27.0 g/mol = 2.70 g. A typical mistake would be dividing (0.100 ÷ 27.0 = 0.00370), which gives the wrong value and units (mol²/g instead of g). Always check your units: when multiplying moles by g/mol, the mol units cancel, leaving grams as desired.